| In this paper we discuss the reconstruction of cracks which are segments in a planar domain .In fact,it is a inverse geometry problem of laplace equation.In the first part,we treat the background.In the second part,we write the mathematical models of the forward problem:Ifσj(j = 1,2…) are insulating cracks:Ifσj(J=1,2…m) are conducting cracks:This two mathematical models can both change Nuemann boundary value condition into the corresponding Dirichlet boundary value condition:u = f,on(?)Ω.The inverse problem is the problem of reconstruction cracks with one or more current flux on (?)Ωand the corresponding voltage potential.Then we prove the duality of conducting cracks and insulating cracks:The solution of a Nuemann boundary value problem of conducting cracks is a harmonic conjugate of a Dirichlet boundary value problem of insulating cracks.In the third part,we first write the asymptotic behaviour near the tip of the crack: Theorem 3.1 u is the solution of a Nuemann boundary value problem of insulating cracks,αis the angle of the crack and the positive x-axis,There are constantsα1,α2,α3…such thatwhere x -x0 = r(cosθ, sinθ).Then by the Green theorem,we prove thatThe functional of the left of (3.56) is called the indicator function.We choose spectral functions -oscillating-decaying functions v(?),ω,t as test functions.where vectorω∈S1 (?) R2,ω⊥is a unit vector perpendicular toω. In the last part,wegive the reconstruction methods of a surface breaking crack, interior crack,polygonalcracks:Reconstruction procedure of a surface breaking crack.step 1 Chooseω1,ω2∈S1. step 2 Computestep 3 Compute h∑(ω2) as same as step 2. step 4 Find tip x0:{ω1·x = h∑(ω1)}∩{ω2·x =h∑(ω2)} = {x0}, (3.59)Reconstruction procedure of a interior crack. step 1 Chooseω1,ω1∈S1. step 2 ComputeCompute h∑(ω1), h∑(-w1),h∑(-ω2) in the same way. step 4 Compute{ω1·x =h∑(w1)}∩{ω2·x = h∑(ω2)} = {x0}. (3.61)If x0∈Ω, x1 = x0 is a crack tip , ture to step 6. step 5 Compute{ω1·x = h∑(ω1)}∩{ω2·x = h∑(-ω2)} = {x0}, (3.62)x1 = x0 is a crack tip , ture to step 7. step 6 Compute{ω1·x = h∑(-ω1)} n {ω2·x = h∑(-ω2)} = {x0}, (3.63)x2 = X0 is a crack tip. step 7 Compute{ω1·x = h∑(-ω1)}∩{ω2·x = h∑(-ω2)} = {x0}, (3.64)x2 = x0 is a another crack tip Reconstruction procedure of polygonal cracks.step 0 Chooseω0∈S1, supposeω0 = (cosθ0,sinθ0);computeIf |t0"-t0'|<ε,stop to compute.step 1 Chooseω1,ω2∈S1. step 2 ComputeCompute h∑(ω1),h∑(-ω1), h∑(-ω2) in the same way. step 4 Compute{ω1·x = h∑(ω1)}∩{ω2·x = h∑(ω2)} = {x0}. (3.61)If x0∈Ω,, x1 = x0 is a crack tip , ture to step 6.step 5 Compute{ω1·x = h∑(ω1)}∩{ω2·x =h∑s(-ω2)} = {x0}, (3.62) = x0 is a crack tip , ture to step 7. step 6 Compute{ω1-x = h∑(-ω1)}∩{ω2·x = h∑(-ω2)} = {x0}, (3.63)x2 = x0 is a crack tip. step 7 Compute{ω1·x = h∑(-ω1)}∩{ω2·x =h∑(-ω2)} = {x0}, (3.64)x3 = x0 is a another crack tipReconstruction procedure of polygonal cracks.step 0 Chooseω0∈S1, supposeω0 = (cosθ0, sinθ0);computeIf |t"0～t'0| <ε,stop to compute. step 1 Suppose v = (cos(θn-1 -π/(3m)),sin(θn-1 -π/(3m))) ,computeIf |t"n - t'n| <ε,stop to compute.step 2 If t"n- t'n< t"n-1-t'n-1,ωn = u,tune to step 1,cycle n. step 3 v = (cos(θn-1) + (π/3m)),sin(θn-1 + (π/3m))) computeIf|t"n—t'n|<ε,stop to compute.step 4 If t"n—t'n < t'n1—t'n1,ωn = u,tune to step 3,cycle n.step 5 Tune to step 1, cycle m.Thus we locate the direction wn of a crack,in the same way we can also locate the direction of another crack .Then we can detection them with the method.of the reconstruction of a interior crack. |
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